Question

A long straight horizontal cable carries a current of 2.5 A in thedirection 10° south of west to 10° north of east. The magnetic meridianof the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dipis zero. Locate the line of neutral points (ignore the thickness of thecable)? (At neutral points, magnetic field due to a current-carryingcable is equal and opposite to the horizontal component of earth’smagnetic field.)

Answer 1

Given, the current in the wire is 2.5 A, the angle of dip is 0° and the magnitude of the Earth’s magnetic field is 0.33 G.

The formula to calculate the horizontal component of the Earth’s magnetic field is,

H H =Hcosδ

Here, the magnitude of the Earth’s magnetic field is H and the angle of dip is δ.

Substituting the values in the above equation, we get:

H H =( 0.33 G×( 10 −4  T 1 G ) )cos0° =3.33× 10 −4  T

The formula to calculate the magnitude of the magnetic field at a distance r on the axis of the magnet is,

H H = μ 0 2π ( I r ) r= μ 0 I 2π H H

Here, the permeability of the free space is μ 0 , the horizontal component of the magnetic field is H H and the distance from the bar magnet is r.

Substituting the values in the above equation, we get:

r= ( 4π× 10 −7 )( 2.5 ) 2π( 0.33× 10 −4 ) =1.51× 10 −2  m ≈1.5 cm

Thus, the set of neutral points parallel to and above the cable are located at a normal distance of 1.5 cm.